if ABCD is a rectangle in which angle BAC=35 degree . find angel DBC.
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Given:- ABCD is a rectangle
Angle BAC=35°
To find:- Angle DBC
Found:- Let intersection point of AC and BD be O.
OB=OC [Diagonals of a rectangle are equal and bisect each other]
Therefore, angle OCB=angle OBC -equation 1
In triangle ABC,
BAC+ABC+ACB=180° [Angle sum property]
35°+90°+ACB=180°
=125°+ACB=180°
=ACB=180°-125°
=ACB=55°
Therefore, Angle DBC=55° [From equation 1]
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